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3 years ago

Chemistry question! Please explain!

Chemistry

1 answer:

Ilya [14]3 years ago

6 0

There are no numbers provided so I will try to explain the equation. If you want to find the percent error you would do theoretical (the number it’s supposed to be) minus experimental (the number you got in the experiment) over the theoretical value (same as the first one.) You must multiply this by 100 in order to show the actual percentage rather than very low decimals. You want the percent error to be as low as possible because the lower it is the less mistakes happened. Mistakes can include too much/less of solvent, solution, or the glassware wasn’t clean enough, and much much more.

I hope this helps!
Please mark the brainliest if it does! (:

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The harmonic law would suggest that Neptune will take how much time to orbit the Sun than Mercury?

Delicious77 [7]

The harmonic law would suggest that Neptune will take a longer amount of time to orbit the sun than Mercury. That's what I think anyways

8 0

2 years ago

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What is the classification of a solution of naoh with a ph of 8.3?

Maksim231197 [3]

Answer:

<u>Alkaline or basic solution </u>(alkaline and basic means the same)

Explanation:

According to the <em>pH</em>, solutions may be classified as neutral, acidic, or alkaline (basic).

This table shows such classification:

pH classification

7 neutral

> 7 alkaline or basic

< 7 acidic

Thus, since the pH of the solution is 8.3, which is greater than 7, the solution is classified as basic (alkaline).

Additionally, you must learn that pH is a logarithmic scale for the concentration of hydronium ions in the solution.

pH = - log [H₃O⁺]

You can calculate the concentration of hydronium ions using antilogarithm properties:

$pH=-log[H_3O^+]\\ \\ {[H_3O^+]}=10^{-pH}\\ \\ {[H_3O^+]}=10^{-8.3}=0.00000000501$

NaOH solutions are alkaline solutions, bases, according to Arrhenius model, because they contain OH⁻ ions and release them when ionize in water.

8 0

3 years ago

How do minerals affect igneous rock formations?

wariber [46]

Mineral composition affects the classification of igneous rock. in simplified classification, igneous to is are classified by the type of feldspar present, by the I type a of feldspar present, or the absence of quartz. in case of neither present, then by the type of iron or magnesium present.

3 0

3 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

2 years ago

Sound _____.

AysviL [449]

Answer:

Sounds travels in transverse waves requires a medium to travel through

4 0

2 years ago

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